How many liters of a .88 M solution can be made with 25.5 grams of lithium fluoride ( $L i F$ $LiF$ )?

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Answer 1 · 2016-05-16T15:25:34

$L = 1.10 L$ $L = 1.10 L$ of solution

The Molarity $M$ $M$ is calculated by the equation comparing moles of solute to liters of solution

$M = \frac{m o l}{L}$ $M=(mol)/L$

For this question we are given the Molarity 0.88M

We are told the solute is a 25.2 gram sample of LiF, Lithium Fluoride

We can convert the mass of LiF to moles by dividing by the molar mass of LiF

Li = 6.94
F = 19.0

LiF = 25.94 g/mole

$25.2 g r a m s x \frac{1 m o l}{25.94 g r a m s}$ $25.2 cancel(grams) x (1 mol)/(25.94cancel(grams))$ = $0.97$ $0.97$ moles

Now we can take the the molarity and the moles and calculate the Liters of solution

$M = \frac{m o l}{L}$ $M=(mol)/L$

$M L = m o l$ $ML= mol$

$L = \frac{m o l}{M}$ $L = (mol)/M$

$L = \frac{0.97 m o l}{0.88 M}$ $L = (0.97mol)/(0.88M)$

$L = 1.10 L$ $L = 1.10 L$ of solution

How many liters of a .88 M solution can be made with 25.5 grams of lithium fluoride (LiFLiF)?